Question

What $\left(H_3{O}^{+}\right)$ must be maintained in a saturated $H_{2}S$ solution to precipitate ${Pb}^{2+}$, but not ${Zn}^{2+}$ from a solution in which each ion is present at a concentration of $0.01M$?
($K_{sp}{H}_{2}S=1.1\times {10}^{-22}$ and $K_{sp}ZnS=1.0\times {10}^{-21}$)

Answer 1

$K_{sp}ZnS=\left[{Zn}^{2+}\right]\left[S^{2-}\right]=1.0\times {10}^{-21}$
or $0.01\times \left[S^{2-}\right]=1.0\times {10}^{-21}$
or $\left[S^{2-}\right]=1.0\times {10}^{-19}$
To start precipitation of $ZnS$, the required concentration of sulphide ions is $1.0\times {10}^{-19}$
$K_{sp}{H}_{2}S={\left[H^{+}\right]}^2\left[S^{2-}\right]=1.1\times {10}^{-22}$
or ${\left[H^{+}\right]}^2=\frac{1.1\times {10}^{-22}}{1.0\times {10}^{-19}}=1.1\times {10}^{-3}$
or $\left[H^{+}\right]=0.0331M$
Thus, the hydrogen ion concentration higher than $0.0331M$ will reduce the concentration of sulphide ions which will not allow the precipitation of ${Zn}^{2+}$ ions as $ZnS$ and under this condition only $PbS$ precipitation occurs.